-0.01d^2+2d+1=0

Solution for -0.01d^2+2d+1=0 equation:

-0.01d^2+2d+1=0

a = -0.01; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-0.01)·1
Δ = 4.04
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-\sqrt{4.04}}{2*-0.01}=\frac{-2-\sqrt{4.04}}{-0.02}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+\sqrt{4.04}}{2*-0.01}=\frac{-2+\sqrt{4.04}}{-0.02}
$